Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$$
I was able to solve the first one:
$\sqrt{n^8+1}\le \sqrt{n^8+k}\le \sqrt{n^8+n}$
$\implies \displaystyle\dfrac{1}{\sqrt{n^8+1}}\ge \dfrac{1}{\sqrt{n^8+k}}\ge \dfrac{1}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{k^3}{\sqrt{n^8+1}}\ge \dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{n^2(n+1)^2}{4\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{n^2(n+1)^2}{4\sqrt{n^8+n}}$
The upper fraction goes to $\frac{1}{4}$ and the lower as well. From the Squeeze theorem:
$$\lim\limits_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}} = \frac{1}{4}$$
The same method did not work for the second limit. Can I get a clue or a hint, please?
 A: There is another way which would allow to answer both questions at the same time using the binomial theorem
$$\frac{k^3}{\sqrt{n^8+k}}=\sum_{n=3}^\infty (n-3)^{20-8 n} \binom{-\frac{1}{2}}{n-3} k ^n=(n-3)^{20-8 n} \binom{-\frac{1}{2}}{n-3}\sum_{n=3}^\infty k^n$$ Now, use Faulhaber's formulae  and you should arrive to something like
$$S_n=\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}=\frac{1}{4}+\frac{1}{2 n}+\frac{1}{4 n^2}+O\left(\frac{1}{n^7}\right)$$
Trying for $n=10$; the summation evaluates to $0.302499987$ while the above truncates expansion gives     $\frac{121}{400}=0.302500000$.
A: The second limit can be squeezed as well with :
$$n^4 \leq \sqrt{n^8+k} \leq n^4+1$$
Thus
$$n\left(\sum_{k=1}^n \frac{k^3}{n^4+1}-\frac{1}{4}\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{n^4}-\frac{1}{4}\right)$$
or
$$n\left[\frac{n^2(n+1)^2}{4(n^4+1)}-\frac{1}{4}\right]\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\left[\frac{n^2(n+1)^2}{4n^4}-\frac{1}{4}\right]$$
or
$$n\cdot \frac{2n^3+n^2-1}{4(n^4+1)}\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)\leq n\cdot \frac{2n^3+n^2}{4n^4}$$
and squeezing
$$\lim_{n\to \infty} n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\right)=\frac{1}{2}$$
A: Note first that $\sum_{k=1}^n k^3=\frac14n^2(n+1)^2$. Now note that $\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+k/n^8}}$ and hence $$\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}\leq\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}\leq\frac1n\sum_{k=1}^n (k/n)^3$$ implying that $\lim_{n\to\infty}\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac14$.
For part 2 again see that
$$n\left(\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}-\frac14\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac14\right)\leq n\left(\frac1n\sum_{k=1}^n (k/n)^3-\frac14\right)$$ which directly implies the second limit your are after is $\frac12$.
